Solution of quadratic equations and Nature of roots Questions in English

(B) Given the polynomial $f(x) = x^3 + a x^2 + b x + c$.
Let the roots of the polynomial be $s - t$,$s$,and $s + t$,where $s$ and $t$ are integers.
According to Vieta's formulas,the sum of the roots is given by:
$(s - t) + s + (s + t) = -a$.
Simplifying this,we get $3s = -a$,or $a = -3s$.
Since $s$ is an integer,$a$ must be a multiple of $3$.
Checking the given options:
$A: -642 = 3 \times (-214)$ (Multiple of $3$)
$B: 1214$ (Not a multiple of $3$,as $1+2+1+4 = 8$)
$C: 1323 = 3 \times 441$ (Multiple of $3$)
$D: 1626 = 3 \times 542$ (Multiple of $3$)
Therefore,'$a$' cannot take the value $1214$.

(B) Let $x = 4+\frac{1}{4+\frac{1}{4+\frac{1}{4+\ldots \infty}}}$.
Since the expression is infinite,we can write $x = 4 + \frac{1}{x}$.
Multiplying by $x$,we get $x^2 = 4x + 1$,which simplifies to $x^2 - 4x - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$.
Since the expression $4+\frac{1}{4+\ldots}$ must be positive,we discard $2-\sqrt{5} \approx -0.236$.
Thus,$x = 2+\sqrt{5}$.

(A) Let $x$ and $x+2$ be two consecutive positive even integers.
Given,$x^2 + (x+2)^2 = 290$.
Expanding the equation: $x^2 + x^2 + 4x + 4 = 290$.
$\Rightarrow 2x^2 + 4x - 286 = 0$.
Dividing by $2$: $x^2 + 2x - 143 = 0$.
Factoring the quadratic equation: $(x + 13)(x - 11) = 0$.
So,$x = -13$ or $x = 11$.
Since $x$ must be a positive even integer,we discard $x = -13$.
For $x = 11$,the consecutive even integer is $x+2 = 13$,which is not even.
Therefore,there are no such pairs of consecutive positive even integers.
The number of solutions is $0$.

(D) Given equation: $4+6(e^{2x}+1) \tanh x = 11 \cosh x + 11 \sinh x$
Substitute $\tanh x = \frac{e^x-e^{-x}}{e^x+e^{-x}}$,$\cosh x = \frac{e^x+e^{-x}}{2}$,and $\sinh x = \frac{e^x-e^{-x}}{2}$:
$4+6(e^{2x}+1) \left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right) = 11 \left(\frac{e^x+e^{-x}}{2}\right) + 11 \left(\frac{e^x-e^{-x}}{2}\right)$
$4+6(e^{2x}+1) \left(\frac{e^{2x}-1}{e^{2x}+1}\right) = 11 \left(\frac{2e^x}{2}\right)$
$4+6(e^{2x}-1) = 11e^x$
$6e^{2x} - 11e^x - 2 = 0$
Let $e^x = t$,where $t > 0$:
$6t^2 - 11t - 2 = 0$
$(6t+1)(t-2) = 0$
$t = 2$ or $t = -\frac{1}{6}$
Since $t > 0$,we have $e^x = 2$,which implies $x = \log_e 2$.

(C) Given the equation $2 \cosh^2 x - 3 \sinh x - k = 0$.
Using the identity $\cosh^2 x = 1 + \sinh^2 x$,we substitute this into the equation:
$2(1 + \sinh^2 x) - 3 \sinh x - k = 0$
$2 \sinh^2 x - 3 \sinh x + (2 - k) = 0$.
Let $t = \sinh x$,where $t \in R$. The quadratic equation $2t^2 - 3t + (2 - k) = 0$ has no real solution for $x$ if the discriminant $D < 0$.
$D = (-3)^2 - 4(2)(2 - k) < 0$
$9 - 8(2 - k) < 0$
$9 - 16 + 8k < 0$
$8k - 7 < 0$
$k < \frac{7}{8}$.

(C) The given equations are $x^2+5x-6=0$ and $y^2-8y-20=0$.
Solving for $x$:
$x^2+6x-x-6=0$ $\Rightarrow (x+6)(x-1)=0$ $\Rightarrow x_1=-6, x_2=1$.
Solving for $y$:
$y^2-10y+2y-20=0$ $\Rightarrow (y-10)(y+2)=0$ $\Rightarrow y_1=10, y_2=-2$.
The vertices of the rectangle are formed by the intersection of these lines: $(-6, 10), (1, 10), (1, -2), (-6, -2)$.
The midpoint of the diagonal is the midpoint of the segment connecting opposite vertices,such as $(-6, 10)$ and $(1, -2)$.
Midpoint $= \left(\frac{-6+1}{2}, \frac{10-2}{2}\right) = \left(\frac{-5}{2}, \frac{8}{2}\right) = \left(\frac{-5}{2}, 4\right)$.

(C) Using the remainder theorem,the remainder is $P(2)$.
Given $P(x) = 2x^3 - 5x^2 + 7$.
Substituting $x = 2$ into the polynomial:
$P(2) = 2(2)^3 - 5(2)^2 + 7$
$P(2) = 2(8) - 5(4) + 7$
$P(2) = 16 - 20 + 7$
$P(2) = 3$
Therefore,the remainder is $3$.

(B) Given that $x^4+x^2+1$ is divisible by both $x^2+mx+1$ and $x^2+nx+1$.
Since the polynomial is of degree $4$ and the divisors are of degree $2$,we can write:
$x^4+x^2+1 = (x^2+mx+1)(x^2+nx+1)$
Expanding the right side:
$x^4 + nx^3 + x^2 + mx^3 + mnx^2 + mx + x^2 + nx + 1$
$= x^4 + (m+n)x^3 + (mn+2)x^2 + (m+n)x + 1$
Comparing the coefficients of $x^3$ on both sides,we get:
$m+n = 0$

(B) Given equations are:
$1) x+y+z=1$
$2) x^2+y^2+z^2=1$
$3) x^3+y^3+z^3=1$
From $(1)$,we have $z = 1 - x - y$. Substituting this into $(2)$:
$x^2 + y^2 + (1 - x - y)^2 = 1$
$x^2 + y^2 + 1 + x^2 + y^2 - 2x - 2y + 2xy = 1$
$2x^2 + 2y^2 + 2xy - 2x - 2y = 0$
$x^2 + y^2 + xy - x - y = 0$
Multiplying by $2$: $2x^2 + 2y^2 + 2xy - 2x - 2y = 0$,which can be written as $(x+y)^2 + x^2 + y^2 - 2x - 2y = 0$.
Alternatively,note that if two variables are $0$,the third must be $1$. For example,if $x=1, y=0, z=0$,then $1+0+0=1$,$1^2+0^2+0^2=1$,and $1^3+0^3+0^3=1$. These satisfy all equations.
Testing permutations of $(1, 0, 0)$,we get $(1, 0, 0)$,$(0, 1, 0)$,and $(0, 0, 1)$.
Thus,there are $3$ solutions.

(B) Given equation is $\frac{(x^2+1)^3}{x^3} + \frac{x^2+1}{3x} = 0$ for $x \neq 0$.
Let $t = \frac{x^2+1}{x}$. Then the equation becomes $t^3 + \frac{t}{3} = 0$.
Factoring out $t$,we get $t(t^2 + \frac{1}{3}) = 0$.
This implies $t = 0$ or $t^2 = -\frac{1}{3}$.
Since $t$ must be a real number,$t^2 = -\frac{1}{3}$ has no real solutions.
Thus,we must have $t = 0$,which means $\frac{x^2+1}{x} = 0$.
This implies $x^2 + 1 = 0$,or $x^2 = -1$.
Since $x^2 = -1$ has no real solutions for $x$,there are no real roots for the given equation.
Therefore,the number of real roots is $0$.

(C) Given equation is $3^{x^2-x} = 25 - 4^{x^2-x}$.
Rearranging the terms,we get $4^{x^2-x} + 3^{x^2-x} = 25$.
We know that $25 = 16 + 9 = 4^2 + 3^2$.
So,the equation becomes $4^{x^2-x} + 3^{x^2-x} = 4^2 + 3^2$.
Comparing the powers,we get $x^2 - x = 2$.
This simplifies to the quadratic equation $x^2 - x - 2 = 0$.
Factoring the quadratic,we get $(x - 2)(x + 1) = 0$.
Thus,the solutions are $x = 2$ and $x = -1$.
Therefore,option $C$ is correct.

(A) To determine if the polynomial $f(x) = x^2-6x+12$ is reducible over $\mathbb{Q}$,we check for its roots using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Here,$a=1, b=-6, c=12$.
The discriminant is $D = b^2-4ac = (-6)^2 - 4(1)(12) = 36 - 48 = -12$.
Since the discriminant $D < 0$,the roots are complex numbers: $x = \frac{6 \pm \sqrt{-12}}{2} = 3 \pm i\sqrt{3}$.
Because the roots are not in $\mathbb{Q}$,the polynomial cannot be factored into linear factors over $\mathbb{Q}$.
Since it is a quadratic polynomial with no roots in $\mathbb{Q}$,it is irreducible over $\mathbb{Q}$.

(D) Given the equation for $x > 2$:
$\sqrt{x+2} - \sqrt{x-2} = \sqrt{4x-2}$
Squaring both sides,we get:
$(x+2) + (x-2) - 2\sqrt{(x+2)(x-2)} = 4x - 2$
$2x - 2\sqrt{x^2-4} = 4x - 2$
$-2\sqrt{x^2-4} = 2x - 2$
$-\sqrt{x^2-4} = x - 1$
Squaring again:
$x^2 - 4 = (x-1)^2$
$x^2 - 4 = x^2 - 2x + 1$
$-4 = -2x + 1$
$2x = 5$
$x = 2.5$
Now,check the value $x = 2.5$ in the original equation:
$LHS$: $\sqrt{2.5+2} - \sqrt{2.5-2} = \sqrt{4.5} - \sqrt{0.5} = \sqrt{9 \times 0.5} - \sqrt{0.5} = 3\sqrt{0.5} - \sqrt{0.5} = 2\sqrt{0.5} = \sqrt{4 \times 0.5} = \sqrt{2}$
$RHS$: $\sqrt{4(2.5) - 2} = \sqrt{10 - 2} = \sqrt{8} = 2\sqrt{2}$
Since $LHS$ $\neq$ $RHS$,there is no solution. Thus,option $D$ is correct.

(B) polynomial $f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$ is defined as a monic polynomial if the leading coefficient,which is the coefficient of the highest degree term $x^n$,is equal to $1$.
Therefore,for the given polynomial $f(x)$ to be monic,we must have $a_n = 1$.

(C) Given,$f(x) = (x - a)(x - b) - (\frac{a + b}{2})$.
Expanding this,we get $f(x) = x^2 - (a + b)x + ab - (\frac{a + b}{2})$.
The minimum value of a quadratic $Ax^2 + Bx + C$ is given by $-\frac{D}{4A}$,where $D = B^2 - 4AC$.
Here,$A = 1$,$B = -(a + b)$,and $C = ab - \frac{a + b}{2}$.
$D = (-(a + b))^2 - 4(1)(ab - \frac{a + b}{2}) = (a + b)^2 - 4ab + 2(a + b) = (a - b)^2 + 2(a + b)$.
Minimum value $= -\frac{(a - b)^2 + 2(a + b)}{4}$.
Since the roots are non-negative,the sum of roots $\alpha + \beta = (a + b) \geq 0$ and the product of roots $\alpha \beta = ab - \frac{a + b}{2} \geq 0$.
Also,for real roots,$D \geq 0$,which is satisfied as $(a - b)^2 + 2(a + b) \geq 0$.
Given the condition of non-negative roots,the minimum value of the function is attained at $x = \frac{a + b}{2}$.
Substituting $x = \frac{a + b}{2}$ into $f(x)$:
$f(\frac{a + b}{2}) = (\frac{a + b}{2} - a)(\frac{a + b}{2} - b) - \frac{a + b}{2} = (\frac{b - a}{2})(\frac{a - b}{2}) - \frac{a + b}{2} = -\frac{(a - b)^2}{4} - \frac{a + b}{2}$.
Given the constraints,the minimum value is $\geq -\frac{(a + b)^2}{4}$.

(C) For the quadratic equation $ax^2 + ax + 5 = 0$ to have no real roots,its discriminant $D$ must be less than $0$.
$D = b^2 - 4ac < 0$
Here,$a = a$,$b = a$,and $c = 5$.
So,$a^2 - 4(a)(5) < 0$
$a^2 - 20a < 0$
$a(a - 20) < 0$
This inequality holds when $0 < a < 20$.
The integral values of '$a$' are $1, 2, 3, \dots, 19$.
The total number of such integral values is $19$.

(C) For the quadratic equation $x^2-3ax+a^2-2a-K=0$ to have distinct real roots,the discriminant $D$ must be greater than $0$.
$D = b^2-4ac > 0$
$(-3a)^2 - 4(1)(a^2-2a-K) > 0$
$9a^2 - 4a^2 + 8a + 4K > 0$
$5a^2 + 8a + 4K > 0$
This inequality must hold for all rational numbers $a$. Since $5a^2 + 8a + 4K$ is a quadratic in $a$ with a positive leading coefficient $(5 > 0)$,it will be positive for all $a$ if its own discriminant $D_a < 0$.
$D_a = (8)^2 - 4(5)(4K) < 0$
$64 - 80K < 0$
$64 < 80K$
$K > \frac{64}{80}$
$K > \frac{4}{5}$
Thus,$K$ lies in the interval $(\frac{4}{5}, \infty)$.

(B) Since $f(x)$ is a second degree polynomial and $f(-3) = 0$,we can write $f(x) = a(x + 3)(x - k)$ for some constant $a$ and root $k$.
Given $f(x) \geq 0$ for all $x \in R$,the graph of $f(x)$ must touch the $x$-axis at exactly one point,meaning it has a double root at $x = -3$. Thus,$k = -3$.
So,$f(x) = a(x + 3)^2$.
Using the condition $f(0) = 18$:
$a(0 + 3)^2 = 18$
$9a = 18$
$a = 2$.
Therefore,$f(x) = 2(x + 3)^2$.
Now,calculating $f(3)$:
$f(3) = 2(3 + 3)^2 = 2(6)^2 = 2 \times 36 = 72$.

(A) Given the polynomial equation $x^5+4 x^4-13 x^3-52 x^2+36 x+144=0$.
We are given that $2$ is one of the roots.
By performing polynomial division or testing factors,we can factor the expression.
Dividing $x^5+4 x^4-13 x^3-52 x^2+36 x+144$ by $(x-2)$,we get the quotient $x^4+6x^3-x^2-54x-72$.
Further factoring,we find the roots of the equation are $-4, -3, -2, 2, 3$.
Given the condition $\alpha < \beta < \gamma < 2 < \varepsilon$,we assign the roots as follows:
$\alpha = -4, \beta = -3, \gamma = -2, \varepsilon = 3$.
Now,calculate the expression:
$\alpha+2 \beta+3 \gamma+5 \varepsilon = (-4) + 2(-3) + 3(-2) + 5(3)$
$= -4 - 6 - 6 + 15$
$= -16 + 15 = -1$.

(C) We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.
$\sin ^2 18^{\circ} = \left(\frac{\sqrt{5}-1}{4}\right)^2 = \frac{5+1-2\sqrt{5}}{16} = \frac{6-2\sqrt{5}}{16} = \frac{3-\sqrt{5}}{8}$.
$\cos ^2 36^{\circ} = \left(\frac{\sqrt{5}+1}{4}\right)^2 = \frac{5+1+2\sqrt{5}}{16} = \frac{6+2\sqrt{5}}{16} = \frac{3+\sqrt{5}}{8}$.
The sum of the roots is $\frac{3-\sqrt{5}}{8} + \frac{3+\sqrt{5}}{8} = \frac{6}{8} = \frac{3}{4}$.
The product of the roots is $\left(\frac{3-\sqrt{5}}{8}\right) \left(\frac{3+\sqrt{5}}{8}\right) = \frac{9-5}{64} = \frac{4}{64} = \frac{1}{16}$.
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
$x^2 - \frac{3}{4}x + \frac{1}{16} = 0$.
Multiplying by $16$,we get $16x^2 - 12x + 1 = 0$.

(D) The equation $f(x)^{g(x)}=1$ is satisfied in the following cases:
Case $1$: $f(x)=1$
$x^2-7x+11=1 \implies x^2-7x+10=0 \implies (x-2)(x-5)=0 \implies x=2, 5$.
Case $2$: $g(x)=0$ and $f(x) \neq 0$
$x^2-6x-7=0 \implies (x-7)(x+1)=0 \implies x=7, -1$.
Case $3$: $f(x)=-1$ and $g(x)$ is an even integer
$x^2-7x+11=-1 \implies x^2-7x+12=0 \implies (x-3)(x-4)=0 \implies x=3, 4$.
Check $g(x)$ for $x=3, 4$:
For $x=3$,$g(3)=3^2-6(3)-7=9-18-7=-16$ (even,so $x=3$ is a solution).
For $x=4$,$g(4)=4^2-6(4)-7=16-24-7=-15$ (odd,so $x=4$ is not a solution).
The set of real values of $x$ is $\{2, 5, 7, -1, 3\}$.
The sum of these values is $2+5+7-1+3=16$.

(C) The given equation is $6x^6-25x^5+31x^4-31x^2+25x-6=0$.
This is a reciprocal equation of the first type.
Dividing by $x^3$,we get $6(x^3 - \frac{1}{x^3}) - 25(x^2 + \frac{1}{x^2}) + 31(x - \frac{1}{x}) = 0$.
Let $t = x - \frac{1}{x}$. Then $x^2 + \frac{1}{x^2} = t^2 + 2$ and $x^3 - \frac{1}{x^3} = t^3 + 3t$.
Substituting these into the equation: $6(t^3 + 3t) - 25(t^2 + 2) + 31t = 0$.
$6t^3 - 25t^2 + 49t - 50 = 0$.
By testing rational roots,$t=2$ is a root: $6(8) - 25(4) + 49(2) - 50 = 48 - 100 + 98 - 50 = -4 \neq 0$.
Actually,checking $t=2$: $6(8) - 25(4) + 49(2) - 50 = 48 - 100 + 98 - 50 = -4$.
Let's re-examine the equation: $6(x^6-1) - 25x(x^4-1) + 31x^2(x^2-1) = 0$.
$(x^2-1)[6(x^4+x^2+1) - 25x(x^2+1) + 31x^2] = 0$.
$(x-1)(x+1)(6x^4 - 25x^3 + 37x^2 - 25x + 6) = 0$.
The roots $x=1$ and $x=-1$ are rational.
For $6x^4 - 25x^3 + 37x^2 - 25x + 6 = 0$,divide by $x^2$: $6(x^2 + \frac{1}{x^2}) - 25(x + \frac{1}{x}) + 37 = 0$.
Let $u = x + \frac{1}{x}$. Then $6(u^2-2) - 25u + 37 = 0 \implies 6u^2 - 25u + 25 = 0$.
$(2u-5)(3u-5) = 0$,so $u = \frac{5}{2}$ or $u = \frac{5}{3}$.
If $x + \frac{1}{x} = \frac{5}{2}$,$2x^2 - 5x + 2 = 0 \implies (2x-1)(x-2) = 0$,so $x = 2, \frac{1}{2}$.
If $x + \frac{1}{x} = \frac{5}{3}$,$3x^2 - 5x + 3 = 0$,which has no real roots.
The rational roots are $\{-1, 1, \frac{1}{2}, 2\}$.
The smallest is $m = -1$ and the greatest is $M = 2$.
Thus,$M-m = 2 - (-1) = 3$.

(C) Given the biquadratic equation $x^4-9x^3+31x^2-49x+30=0$.
Since the coefficients are real,complex roots occur in conjugate pairs. Thus,if $(2-i)$ is a root,then $(2+i)$ is also a root.
Let the four roots be $\alpha, \beta, (2-i),$ and $(2+i)$.
The sum of the roots is $\alpha + \beta + (2-i) + (2+i) = -(-9)/1 = 9$.
$\alpha + \beta + 4 = 9 \implies \alpha + \beta = 5$.
The product of the roots is $\alpha \cdot \beta \cdot (2-i)(2+i) = 30/1 = 30$.
Since $(2-i)(2+i) = 2^2 - i^2 = 4 + 1 = 5$,we have $\alpha \cdot \beta \cdot 5 = 30 \implies \alpha \cdot \beta = 6$.
We have $\alpha + \beta = 5$ and $\alpha \cdot \beta = 6$. Solving these,we get $\alpha = 2$ and $\beta = 3$ (since $\alpha < \beta$).
Therefore,$2\alpha - \beta = 2(2) - 3 = 4 - 3 = 1$.

(C) Given the equation $3p^2x^3 + px^2 + qx + 3 = 0$. Substituting $p = 1$ and $q = -7$,we get:
$3x^3 + x^2 - 7x + 3 = 0$.
By testing $x = 1$,we find $3(1)^3 + (1)^2 - 7(1) + 3 = 3 + 1 - 7 + 3 = 0$.
Thus,$(x - 1)$ is a factor. Dividing the polynomial by $(x - 1)$,we get:
$(x - 1)(3x^2 + 4x - 3) = 0$.
The roots are $x = 1$ and the roots of $3x^2 + 4x - 3 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:
$x = \frac{-4 \pm \sqrt{16 - 4(3)(-3)}}{2(3)} = \frac{-4 \pm \sqrt{16 + 36}}{6} = \frac{-4 \pm \sqrt{52}}{6} = \frac{-4 \pm 2\sqrt{13}}{6} = \frac{-2 \pm \sqrt{13}}{3}$.
Let $\alpha = \frac{-2 + \sqrt{13}}{3}$ and $\beta = \frac{-2 - \sqrt{13}}{3}$.
Then $|\alpha - \beta| = |\frac{-2 + \sqrt{13} - (-2 - \sqrt{13})}{3}| = |\frac{2\sqrt{13}}{3}| = \frac{2\sqrt{13}}{3}$.

(C) Given the cubic equation $x^3-3x^2-4x+12=0$.
Factoring the equation:
$x^2(x-3)-4(x-3)=0$
$(x^2-4)(x-3)=0$
$(x-2)(x+2)(x-3)=0$
Thus,the roots are $\alpha=-2, \beta=2, \gamma=3$.
We need to calculate $\sum(\alpha+\beta)^2 = (\alpha+\beta)^2 + (\beta+\gamma)^2 + (\gamma+\alpha)^2$.
Substituting the values:
$(\alpha+\beta)^2 = (-2+2)^2 = 0^2 = 0$
$(\beta+\gamma)^2 = (2+3)^2 = 5^2 = 25$
$(\gamma+\alpha)^2 = (3-2)^2 = 1^2 = 1$
Summing these values: $0 + 25 + 1 = 26$.

(C) Given that the quadratic equation $ax^2+bx+c=0$ has imaginary roots,the discriminant $D = b^2 - 4ac < 0$,which implies $b^2 < 4ac$.
Let $f(x) = 3a^2x^2 + 6abx + 2b^2$.
Since the coefficient of $x^2$ is $3a^2 > 0$,the expression has a minimum value.
The minimum value of a quadratic $Ax^2 + Bx + C$ is given by $\frac{4AC - B^2}{4A}$.
Here,$A = 3a^2$,$B = 6ab$,and $C = 2b^2$.
Minimum value $= \frac{4(3a^2)(2b^2) - (6ab)^2}{4(3a^2)} = \frac{24a^2b^2 - 36a^2b^2}{12a^2} = \frac{-12a^2b^2}{12a^2} = -b^2$.
Since $b^2 < 4ac$,multiplying by $-1$ reverses the inequality: $-b^2 > -4ac$.
Thus,the minimum value is greater than $-4ac$.

(A) Let the roots of the equation $x^4+x^3-4x^2+x+1=0$ be diminished by $h$. Substituting $x$ with $x+h$,the equation becomes $(x+h)^4+(x+h)^3-4(x+h)^2+(x+h)+1=0$.
Expanding the terms,the coefficient of $x^2$ is given by $6h^2+3h-4$.
Since the $x^2$ term is absent,we set $6h^2+3h-4=0$.
The roots of this quadratic equation are $\alpha$ and $\beta$.
Using the relation for the difference of roots,$(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$.
From the quadratic equation $6h^2+3h-4=0$,we have $\alpha+\beta = -\frac{3}{6} = -\frac{1}{2}$ and $\alpha\beta = -\frac{4}{6} = -\frac{2}{3}$.
Thus,$(\alpha-\beta)^2 = (-\frac{1}{2})^2 - 4(-\frac{2}{3}) = \frac{1}{4} + \frac{8}{3} = \frac{3+32}{12} = \frac{35}{12}$.
Therefore,$12(\alpha-\beta)^2 = 12 \times \frac{35}{12} = 35$.

(A) For the roots of the quadratic equation $\lambda x^2 + 13x + 7 = 0$ to be rational,the discriminant $D$ must be a perfect square of a rational number. Since $\lambda$ is an integer,$D$ must be a perfect square of an integer.
$D = b^2 - 4ac = (13)^2 - 4(\lambda)(7) = 169 - 28\lambda$.
Given $\lambda \in (-3, 7)$,the possible integer values for $\lambda$ are $\{-2, -1, 0, 1, 2, 3, 4, 5, 6\}$.
We test these values:
If $\lambda = -2$,$D = 169 - 28(-2) = 169 + 56 = 225 = (15)^2$ (Perfect square).
If $\lambda = -1$,$D = 169 - 28(-1) = 197$ (Not a perfect square).
If $\lambda = 0$,the equation becomes $13x + 7 = 0$,which gives $x = -7/13$ (Rational).
If $\lambda = 1$,$D = 169 - 28 = 141$ (Not a perfect square).
If $\lambda = 2$,$D = 169 - 56 = 113$ (Not a perfect square).
If $\lambda = 3$,$D = 169 - 84 = 85$ (Not a perfect square).
If $\lambda = 4$,$D = 169 - 112 = 57$ (Not a perfect square).
If $\lambda = 5$,$D = 169 - 140 = 29$ (Not a perfect square).
If $\lambda = 6$,$D = 169 - 168 = 1 = (1)^2$ (Perfect square).
Thus,the set $S = \{-2, 0, 6\}$.
The sum of the elements in $S$ is $-2 + 0 + 6 = 4$.

(D) Given equation is $x^3+3x^2-x-3=0$.
Factorizing the equation:
$x^2(x+3)-1(x+3)=0$
$(x^2-1)(x+3)=0$
$(x-1)(x+1)(x+3)=0$
So,the roots are $\alpha = -3, \beta = -1, \gamma = 1$.
Now,we calculate the expression:
$(1+\alpha^2)(1+\beta^2)(1+\gamma^2) = (1+(-3)^2)(1+(-1)^2)(1+(1)^2)$
$= (1+9)(1+1)(1+1)$
$= (10)(2)(2) = 40$.

(C) Let $y = x^2 - 2nx$. The equation becomes $(x-n)(y^2 + (2n^2-5)y + (n^4-5n^2+4)) = 0$.
Factoring the quadratic in $y$: $y^2 + (2n^2-5)y + (n^2-1)(n^2-4) = (y + n^2-1)(y + n^2-4) = 0$.
Substituting $y$ back: $(x-n)(x^2 - 2nx + n^2 - 1)(x^2 - 2nx + n^2 - 4) = 0$.
This simplifies to $(x-n)((x-n)^2 - 1)((x-n)^2 - 4) = 0$.
$(x-n)(x-n-1)(x-n+1)(x-n-2)(x-n+2) = 0$.
The roots are $x = n, n+1, n-1, n+2, n-2$.
The product of the roots $P = n(n+1)(n-1)(n+2)(n-2) = (n-2)(n-1)n(n+1)(n+2)$.
This is the product of $5$ consecutive integers,which is always divisible by $5! = 120$.

(A) Given that $2+\sqrt{3}$ is a root of $f(x)=x^4+2x^3-16x^2-22x+7=0$.
Since the coefficients are rational,the conjugate $2-\sqrt{3}$ must also be a root.
Let $x=2+\sqrt{3}$,then $(x-2)^2=3$,which simplifies to $x^2-4x+1=0$.
Dividing $f(x)$ by $x^2-4x+1$,we get the quotient $x^2+6x+7$.
Setting $x^2+6x+7=0$,we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$x = \frac{-6 \pm \sqrt{36-28}}{2} = \frac{-6 \pm \sqrt{8}}{2} = -3 \pm \sqrt{2}$.
The roots of $f(x)=0$ are $2+\sqrt{3}, 2-\sqrt{3}, -3+\sqrt{2}, -3-\sqrt{2}$.
Comparing these with the given options,$3-\sqrt{2}$ is not a root.

(B) We have,$x^2+|x-3|=4$.
Case $I$: $x \ge 3$.
The equation becomes $x^2 + x - 3 = 4$,which simplifies to $x^2 + x - 7 = 0$.
The roots are $x = \frac{-1 \pm \sqrt{1 - 4(1)(-7)}}{2} = \frac{-1 \pm \sqrt{29}}{2}$.
Since $\sqrt{29} \approx 5.38$,the roots are $\approx 2.19$ and $\approx -3.19$.
Neither of these satisfies the condition $x \ge 3$.
Case $II$: $x < 3$.
The equation becomes $x^2 - (x - 3) = 4$,which simplifies to $x^2 - x + 3 = 4$,or $x^2 - x - 1 = 0$.
The roots are $x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$.
Both roots $\frac{1 + \sqrt{5}}{2} \approx 1.618$ and $\frac{1 - \sqrt{5}}{2} \approx -0.618$ satisfy the condition $x < 3$.
The sum of these roots is $\frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2} = \frac{2}{2} = 1$.

(B) Let the roots of the equation $2x^3 + ax^2 - 8x + b = 0$ be $\alpha, \beta, \gamma$.
After reducing each root by $1$,the new roots are $\alpha-1, \beta-1, \gamma-1$.
Let $y = x - 1$,so $x = y + 1$.
Substituting $x = y + 1$ into the original equation:
$2(y+1)^3 + a(y+1)^2 - 8(y+1) + b = 0$
$2(y^3 + 3y^2 + 3y + 1) + a(y^2 + 2y + 1) - 8(y + 1) + b = 0$
$2y^3 + (6 + a)y^2 + (6 + 2a - 8)y + (2 + a - 8 + b) = 0$.
Given that the coefficient of $y^2$ and the constant term vanish:
$6 + a = 0 \Rightarrow a = -6$.
$2 + a - 8 + b = 0$ $\Rightarrow 2 - 6 - 8 + b = 0$ $\Rightarrow b = 12$.
The original equation is $2x^3 - 6x^2 - 8x + 12 = 0$.
Dividing by $2$,we get $x^3 - 3x^2 - 4x + 6 = 0$.
Since $x = 1$ is a root $(1 - 3 - 4 + 6 = 0)$,we divide by $(x - 1)$:
$(x - 1)(x^2 - 2x - 6) = 0$.
The roots are $x = 1$ and $x = \frac{2 \pm \sqrt{4 - 4(1)(-6)}}{2} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}$.
Thus,the roots are $1, 1 + \sqrt{7}, 1 - \sqrt{7}$.

(A) We have,$(x-2)(x-3) = k^2$,where $k \in R$.
Expanding the equation: $x^2 - 5x + 6 - k^2 = 0$.
Comparing this with the standard quadratic equation $ax^2 + bx + c = 0$,we get $a = 1$,$b = -5$,and $c = 6 - k^2$.
The discriminant $D$ is given by $D = b^2 - 4ac$.
Substituting the values: $D = (-5)^2 - 4(1)(6 - k^2) = 25 - 24 + 4k^2 = 1 + 4k^2$.
Since $k^2 \ge 0$ for all $k \in R$,it follows that $1 + 4k^2 \ge 1$.
Thus,$D > 0$.
Since the discriminant is strictly positive,the roots are always real and distinct.

(C) Given the inequality $f(x) = 9mx - 1 + \frac{1}{x} \geq 0$ for all $x > 0$.
Multiplying by $x$ (since $x > 0$),we get $9mx^2 - x + 1 \geq 0$.
For a quadratic expression $ax^2 + bx + c \geq 0$ to hold for all $x > 0$,we analyze the discriminant $D = b^2 - 4ac$.
Here,$a = 9m$,$b = -1$,and $c = 1$.
The discriminant $D = (-1)^2 - 4(9m)(1) = 1 - 36m$.
For the quadratic to be non-negative for all $x$,we require $D \leq 0$.
$1 - 36m \leq 0 \implies 36m \geq 1 \implies m \geq \frac{1}{36}$.
Thus,the smallest value of $m$ is $\frac{1}{36}$.

(C) Given equation: $|x|^2-5|x|+6=0$
Let $|x|=y$. Since $|x| \ge 0$,we have $y \ge 0$.
The equation becomes $y^2-5y+6=0$.
Factoring the quadratic: $(y-2)(y-3)=0$.
This gives $y=2$ or $y=3$.
Substituting back $|x|=y$:
Case $1$: $|x|=2 \Rightarrow x = \pm 2$.
Case $2$: $|x|=3 \Rightarrow x = \pm 3$.
Thus,the real roots are $x \in \{-3, -2, 2, 3\}$.
Therefore,the number of real roots is $4$.

(C) Given $3 \cos A + 2 = 0$,we have $\cos A = -\frac{2}{3}$.
Since $A$ is an angle in a triangle,$\sin A > 0$. Thus,$\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$.
Then,$\tan A = \frac{\sin A}{\cos A} = \frac{\sqrt{5}/3}{-2/3} = -\frac{\sqrt{5}}{2}$.
The quadratic equation with roots $\alpha$ and $\beta$ is $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.
Here,$\alpha = \sin A = \frac{\sqrt{5}}{3}$ and $\beta = \tan A = -\frac{\sqrt{5}}{2}$.
Sum of roots: $\alpha + \beta = \frac{\sqrt{5}}{3} - \frac{\sqrt{5}}{2} = \frac{2\sqrt{5} - 3\sqrt{5}}{6} = -\frac{\sqrt{5}}{6}$.
Product of roots: $\alpha\beta = \left(\frac{\sqrt{5}}{3}\right) \left(-\frac{\sqrt{5}}{2}\right) = -\frac{5}{6}$.
The equation is $x^2 - (-\frac{\sqrt{5}}{6})x - \frac{5}{6} = 0$,which simplifies to $x^2 + \frac{\sqrt{5}}{6}x - \frac{5}{6} = 0$.
Multiplying by $6$,we get $6x^2 + \sqrt{5}x - 5 = 0$.

(C) Given that $\alpha$ and $\beta$ satisfy the equation $x^2-6x+1=0$.
Let $y = \frac{x}{x+1}$.
Then $y(x+1) = x$,which implies $yx + y = x$,or $x(1-y) = y$,so $x = \frac{y}{1-y}$.
Substituting this into the original equation $x^2-6x+1=0$:
$(\frac{y}{1-y})^2 - 6(\frac{y}{1-y}) + 1 = 0$.
Multiplying by $(1-y)^2$,we get:
$y^2 - 6y(1-y) + (1-y)^2 = 0$.
$y^2 - 6y + 6y^2 + 1 - 2y + y^2 = 0$.
$8y^2 - 8y + 1 = 0$.
Replacing $y$ with $x$,the required equation is $8x^2-8x+1=0$.

(C) Let the roots of the quadratic equation $x^2 - px + q = 0$ be $\alpha$ and $\beta$,where $\alpha, \beta \in \mathbb{Z}^+$.
By the properties of roots,we have:
$\alpha + \beta = p$ (sum of roots)
$\alpha \cdot \beta = q$ (product of roots)
Since $q$ is a prime number,its only factors are $1$ and $q$. Thus,the roots must be $1$ and $q$.
Substituting these into the sum of roots equation: $1 + q = p$.
Since $p$ and $q$ are both prime numbers,we look for two primes that differ by $1$. The only such primes are $2$ and $3$ (where $q=2$ and $p=3$).
Substituting $p=3$ and $q=2$ into the equation: $x^2 - 3x + 2 = 0$.
Factoring the quadratic: $(x - 1)(x - 2) = 0$.
Thus,the roots are $1$ and $2$.

(C) Given equation: $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$.
Let $t = x-a$. Then the equation becomes $t(t-1)+(t-1)(t-2)+t(t-2)=0$.
Expanding the terms: $(t^2-t) + (t^2-3t+2) + (t^2-2t) = 0$.
Combining like terms: $3t^2 - 6t + 2 = 0$.
The discriminant $D = b^2 - 4ac = (-6)^2 - 4(3)(2) = 36 - 24 = 12$.
Since $D > 0$,the roots for $t$ are real and distinct.
Consequently,$x = a + t$ will also be real and distinct for any $a \in R$.

(D) The condition for the roots of $ax^2 + bx + c = 0$ to be in the ratio $m:n$ is $mnb^2 = ac(m+n)^2$.
$(i)$ If $\alpha = \beta$,then the ratio is $1:1$. Substituting $m=1, n=1$ into the formula: $(1)(1)b^2 = ac(1+1)^2 \Rightarrow b^2 = 4ac$. This matches $(E)$.
$(ii)$ If $\alpha = 2\beta$,then the ratio is $2:1$. Substituting $m=2, n=1$: $(2)(1)b^2 = ac(2+1)^2 \Rightarrow 2b^2 = 9ac$. This matches $(B)$.
$(iii)$ If $\alpha = 3\beta$,then the ratio is $3:1$. Substituting $m=3, n=1$: $(3)(1)b^2 = ac(3+1)^2 \Rightarrow 3b^2 = 16ac$. This matches $(D)$.
$(iv)$ If $\alpha = \beta^2$,then $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$. Substituting $\alpha = \beta^2$,we get $\beta^2 + \beta = -b/a$ and $\beta^3 = c/a$. Thus $\beta = (c/a)^{1/3}$. Substituting this into the sum equation: $(c/a)^{2/3} + (c/a)^{1/3} = -b/a$. Multiplying by $a$: $a(c/a)^{2/3} + a(c/a)^{1/3} = -b$ $\Rightarrow (a^3 c^2/a^2)^{1/3} + (a^3 c/a)^{1/3} = -b$ $\Rightarrow (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$. This matches $(A)$.
Therefore,the correct match is $i-E, ii-B, iii-D, iv-A$.

(C) Let $f(x) = x^4+4x^3-16x-16$.
To find the multiple roots,we find the derivative $f'(x) = 4x^3+12x^2-16$.
Setting $f'(x) = 0$,we get $4(x^3+3x^2-4) = 0$.
By inspection,$x=1$ is a root of $f'(x)$,so $(x-1)$ is a factor.
Dividing $x^3+3x^2-4$ by $(x-1)$,we get $(x-1)(x^2+4x+4) = (x-1)(x+2)^2 = 0$.
The roots of $f'(x)=0$ are $x=1$ and $x=-2$.
Checking these in $f(x)$:
$f(1) = 1+4-16-16 = -27 \neq 0$.
$f(-2) = (-2)^4+4(-2)^3-16(-2)-16 = 16-32+32-16 = 0$.
Since $f(-2)=0$ and $f'(-2)=0$,$x=-2$ is a multiple root.
For $x=-2$ to be a root of a quadratic equation $x^2+ax+b=0$,we check the options:
$A: (-2)^2+2(-2)-3 = 4-4-3 = -3 \neq 0$.
$B: (-2)^2-3(-2)+2 = 4+6+2 = 12 \neq 0$.
$C: (-2)^2+(-2)-2 = 4-2-2 = 0$.
$D: (-2)^2-4(-2)+3 = 4+8+3 = 15 \neq 0$.
Thus,the equation is $x^2+x-2=0$.

(A) For a quadratic equation $ax^2 + bx + c = 0$ to have real and equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.
Given equation: $3x^2 + (2k + 1)x - 5k = 0$.
Here,$a = 3$,$b = (2k + 1)$,and $c = -5k$.
$D = (2k + 1)^2 - 4(3)(-5k) = 0$.
$4k^2 + 4k + 1 + 60k = 0$.
$4k^2 + 64k + 1 = 0$.
Using the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$k = \frac{-64 \pm \sqrt{64^2 - 4(4)(1)}}{2(4)} = \frac{-64 \pm \sqrt{4096 - 16}}{8} = \frac{-64 \pm \sqrt{4080}}{8}$.
$k = \frac{-64 \pm 4\sqrt{255}}{8} = \frac{-16 \pm \sqrt{255}}{2}$.
Since $-\frac{1}{2} < k < 0$,we check the values: $\sqrt{255} \approx 15.96$.
$k_1 = \frac{-16 + 15.96}{2} \approx -0.02$ (which is in the range $(-\frac{1}{2}, 0)$).
$k_2 = \frac{-16 - 15.96}{2} \approx -15.98$ (which is not in the range).

(C) Let $x_1, x_2, x_3$ be the roots of the equation $E_1: x^3+x^2+lx+n=0$.
From Vieta's formulas,$x_1+x_2+x_3 = -1$.
Given $E_2: x^3+ax^2+bx+c=0$ is a reciprocal equation of class one,so $c=1$ and $a=b$.
Thus,$E_2: x^3+ax^2+ax+1=0$.
The roots of $E_2$ are $\frac{x_i-1}{2}$.
The sum of roots of $E_2$ is $\sum \frac{x_i-1}{2} = \frac{(x_1+x_2+x_3)-3}{2} = \frac{-1-3}{2} = -2$.
From $E_2$,the sum of roots is $-a$,so $-a = -2 \Rightarrow a=2$.
$E_2$ becomes $x^3+2x^2+2x+1 = (x+1)(x^2+x+1) = 0$.
The roots of $E_2$ are $-1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$.
Using $\frac{x_i-1}{2} = y_i$,we have $x_i = 2y_i+1$.
For $y_1 = -1$,$x_1 = 2(-1)+1 = -1$.
For $y_2 = \frac{-1+i\sqrt{3}}{2}$,$x_2 = 2(\frac{-1+i\sqrt{3}}{2})+1 = i\sqrt{3}$.
For $y_3 = \frac{-1-i\sqrt{3}}{2}$,$x_3 = 2(\frac{-1-i\sqrt{3}}{2})+1 = -i\sqrt{3}$.
The roots of $E_1$ are $\{-1, i\sqrt{3}, -i\sqrt{3}\}$.
The roots of $E_2$ are $\{-1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}\}$.
The common root is $-1$.
Excluding the common root,the remaining roots are $\{i\sqrt{3}, -i\sqrt{3}, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}\}$.

(C) Given that $3$ is a root of the equation $x^2+kx-24=0$.
Substituting $x=3$ into the equation:
$(3)^2 + k(3) - 24 = 0$
$9 + 3k - 24 = 0$
$3k - 15 = 0$
$3k = 15 \Rightarrow k = 5$.
Now,we check the options by substituting $k=5$ and $x=3$:
For option $C$: $x^2 - kx + 6 = 0$
Substituting $x=3$ and $k=5$:
$(3)^2 - (5)(3) + 6 = 9 - 15 + 6 = 0$.
Since the equation is satisfied,$3$ is a root of $x^2-kx+6=0$.